Question

A binomial experiment has the given number of trials N and the given success probability P.
N= 16, p=0.2

Determine the probability P(1 or fewer) round to at least 4 decimal places.

Answer 1

Are you sure the answer isnt .1407?

Answer 2

Find the mean and round to 2 decimal places

Answer 3

(c)Find the variance and standard deviation. Round the variance to two decimal places and standard deviation to at least three decimal places.

Answer 4

To determine the probability of getting 1 or fewer successes in a binomial experiment with \( N = 16 \) trials and a success probability of \( p = 0.2 \), we can use the binomial probability formula. The formula for the probability of getting exactly \( k \) successes is:

\[
P(X = k) = \binom{N}{k} p^k (1-p)^{N-k}
\]

Where:
- \( \binom{N}{k} \) is the binomial coefficient, which represents the number of ways to choose \( k \) successes from \( N \) trials.
- \( p \) is the probability of success on each trial.
- \( (1-p) \) is the probability of failure on each trial.

We want to find \( P(X \leq 1) = P(X = 0) + P(X = 1) \).

### Step 1: Calculate \( P(X = 0) \)
\[
P(X = 0) = \binom{16}{0} (0.2)^0 (0.8)^{16}
\]
\[
= 1 \cdot 1 \cdot (0.8)^{16}
\]
\[
= (0.8)^{16}
\]
\[
\approx 0.0407 \quad (\text{using a calculator for } (0.8)^{16})
\]

### Step 2: Calculate \( P(X = 1) \)
\[
P(X = 1) = \binom{16}{1} (0.2)^1 (0.8)^{15}
\]
\[
= 16 \cdot (0.2) \cdot (0.8)^{15}
\]
\[
= 16 \cdot 0.2 \cdot (0.8)^{15}
\]
\[
= 3.2 \cdot (0.8)^{15}
\]
\[
\approx 3.2 \cdot 0.0512 \quad (\text{using a calculator for } (0.8)^{15})
\]
\[
\approx 0.1638
\]

### Step 3: Calculate \( P(X \leq 1) \)
Now we add \( P(X = 0) \) and \( P(X = 1) \):
\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]
\[
\approx 0.0407 + 0.1638 = 0.2045
\]

### Final Answer
Thus, the probability \( P(1 \text{ or fewer successes}) \) is approximately:
\[
\boxed{0.2045}
\]

Answer 5

Let's go through the calculations step-by-step again to ensure accuracy.

We are calculating the probability \( P(X \leq 1) \) for a binomial distribution with:
- \( N = 16 \)
- \( p = 0.2 \)

This means we want to find:
\[
P(X = 0) + P(X = 1)
\]

### Step 1: Calculate \( P(X = 0) \)

Using the binomial probability formula:
\[
P(X = 0) = \binom{16}{0} (0.2)^0 (0.8)^{16}
\]
\[
= 1 \times 1 \times (0.8)^{16} = (0.8)^{16}
\]

Calculating \( (0.8)^{16} \):
\[
(0.8)^{16} \approx 0.02899
\]

### Step 2: Calculate \( P(X = 1) \)

Now, we calculate \( P(X = 1) \):
\[
P(X = 1) = \binom{16}{1} (0.2)^1 (0.8)^{15}
\]
\[
= 16 \times (0.2) \times (0.8)^{15}
\]
Calculating \( (0.8)^{15} \):
\[
(0.8)^{15} \approx 0.03669
\]
Thus,
\[
P(X = 1) = 16 \times 0.2 \times 0.03669 \approx 1.17408 \times 0.03669 \approx 0.07269
\]

### Step 3: Combine the probabilities

Now we combine \( P(X = 0) \) and \( P(X = 1) \):
\[
P(X \leq 1) = P(X = 0) + P(X = 1)
\]
\[
\approx 0.02899 + 0.07269 \approx 0.10168
\]

### Correct Final Result
Thus, the probability \( P(X \leq 1) \) is approximately:
\[
\boxed{0.1017} (rounded to 4 decimal places)
\]

It seems that I made some mistakes in the previous calculations of \( (0.8)^{15} \) and \( P(X = 1) \). Thank you for your patience; I appreciate your understanding while we clarified the calculations! The result is indeed around \( 0.1017 \), and it does not match either \( 0.2045 \) or \( 0.1407 \).

If you have any further questions or calculations you'd like to explore, please let me know!

Answer 6

In a binomial distribution, the mean (expected value) can be calculated using the formula:

\[
\mu = N \cdot p
\]

Where:
- \( N \) is the number of trials
- \( p \) is the probability of success on each trial

Given:
- \( N = 16 \)
- \( p = 0.2 \)

### Step 1: Calculate the Mean

Using the given values:

\[
\mu = 16 \cdot 0.2 = 3.2
\]

### Final Result

Thus, the mean of the binomial distribution, rounded to two decimal places, is:

\[
\boxed{3.20}
\]

Answer 7

In a binomial distribution, the variance and standard deviation can be calculated using the following formulas:

1. **Variance** (\( \sigma^2 \)):
\[
\sigma^2 = N \cdot p \cdot (1 - p)
\]

2. **Standard Deviation** (\( \sigma \)):
\[
\sigma = \sqrt{\sigma^2}
\]

Given:
- \( N = 16 \)
- \( p = 0.2 \)
- \( 1 - p = 0.8 \)

### Step 1: Calculate the Variance

First, we calculate the variance:

\[
\sigma^2 = 16 \cdot 0.2 \cdot 0.8
\]
\[
= 16 \cdot 0.16
\]
\[
= 2.56
\]

### Step 2: Calculate the Standard Deviation

Next, we calculate the standard deviation:

\[
\sigma = \sqrt{2.56} \approx 1.596
\]

### Final Results

- The variance, rounded to two decimal places, is:
\[
\sigma^2 \approx \boxed{2.56}
\]

- The standard deviation, rounded to at least three decimal places, is:
\[
\sigma \approx \boxed{1.596}
\]